Problem: Let $f(x)=2e^x-5\text{cos}(x)+3$. $f'(x)=$
Answer: Recall that ${\dfrac{d}{dx}[e^x]=e^x}$ and ${\dfrac{d}{dx}[\text{cos}(x)]=-\text{sin}(x)}$. $\begin{aligned} f'(x)&=\dfrac{d}{dx}[2e^x-5\text{cos}(x)+3] \\\\ &=2{\dfrac{d}{dx}[e^x]}-5{\dfrac{d}{dx}[\text{cos}(x)]}+\dfrac{d}{dx}[3] \\\\ &=2\cdot {e^x}-5[{-\text{sin}(x)}]+0 \\\\ &=2e^x+5\text{sin}(x) \end{aligned}$ In conclusion, $f'(x)=2e^x+5\text{sin}(x)$